Vectors a1=(1,1,1,1), a2=(1,-1,1,-1), a3 = (1,3,1,3), b1=(1,2,0,2). b2 = (1,2,1,2) and b3=(3,1,3,1) are given in vector space $\displaystyle R^4$. Find dimL({a1,a2,a3,b1,b2,b3}).
I got this a+b+c+d+e+3f=0,a-b+3c+2d+2e+f=0,a+b+c&# 43;e+3f=0,a-b+3c+2d+2*e+1*f=0 - Wolfram|Alpha so does that means that vectors b2 = (1,2,1,2) and b3=(3,1,3,1) are maximally linearly independent and dimension of L({a1,a2,a3,b1,b2,b3}) is 2?
Oh, dear, do we have to let machines do everything for us? Aren't we losing track of understanding what's going on?
The problem really is to determine if the vectors a1=(1,1,1,1), a2=(1,-1,1,-1), a3 = (1,3,1,3), b1=(1,2,0,2). b2 = (1,2,1,2) and b3=(3,1,3,1) are independent and, if not, what subset of them is independent. From the definition of "independent", we look for numbers, r, s, t, u, v, w, such that ra1+ sa2+ ta3+ ub1+ vb2+ wb3= r(1,1,1,1)+ s(1, -1, 1, -1)+ t(1, 3, 1, 3)+ u(1, 2, 0, 2)+ v(1, 2, 1, 2)+ w(3, 1, 3, 1)= (r+ s+ t+ u+ v+ w, r- s+ 3t+ 2u+ 2v+ 3w, r+ s+ t+ v+ 3w, r- s+ 3t+ 2u+ 2v+ w)= (0, 0, 0, 0). That gives 4 equations in 6 unknowns: r+ s+ t+ u+ v+ w= 0, r- s + 3t+ 2u+ 2v+ 3w= 0, r+ s+ t+ v+ 3w= 0, and r- s+ 3t+ 2u+ 2v+ w= 0. One obvious solution is r= s= t= u= v= w= 0 but question is "are there other, non-trivial, solutions?" Because there are more unknowns than equations, more vectors than the dimension of the space, of course, there are. The question is how many more.
Adding the first two equations eliminates t: 2r+ 4t+ 3u+ 3v+ 4w= 0. Adding the last two does the same: 2r+ 4t+ 2u+ 3v+ 4w= 0. Subtracting the second of those from the first eliminates r, t, v and w, leaving only 3u- 2u= 0 so u= 0. Setting u= 0, both of those equations say that 2r+ 4t+ 3v+ 4w= 0. We can solve that for r as a functio of the other three: r= -2t- (3/2)v- 2w. Putting that, and u= 0, into the very first equation, r+ s+ t+ u+ v+ w= -2t- (3/2)v-2w+ s+ t+ 0+ v+ w= s -t- (1/2)v- w= 0. We can solve that for s: s= t+ (1/2)v+ w. Putting all three of those into the third of our original equation, r+ s+ t+ v+ 3w= -2t- (3/2)v- 2w+ t+ (1/2)v+ w+ t+ v + 3w= -t- v- w= 0 so t= -v- w. Since we got our second set of equations from combining the first and second equations and combining the third and fourth, putting those into the second and fourth equations won't give anything new.
We now know that we can write all of the coeffients in terms of v and w which means we could, say, take v= 1, w= 0 and write the vector (1, 2, 1, 2) (the vector multiplied by v in our linear combination) in terms of the other four. Or we could take v= 0, w= 1 and write (3, 1, 3, 1) in terms of the other four. We don't need those two vectors since they can be written in terms of the other four, but we do need those four. The dimension is 6- 2= 4, not 2.